Question

Today Van de Graaff accelerators sometimes serve as “injectors” for other types of accelerators that then further increase the energy of the particles. Consider a Van de Graaff accelerator that is being used to accelerate protons. The high voltage terminal (metal sphere) of the Van de Graff is charged using a rubberized belt that is 30 cm wide and travels at a velocity of 20 m/s. Charge is sprayed onto the belt near the roller at the low voltage end and removed from the belt near the upper roller inside the high voltage terminal. The belt is given sufficient surface charge density at the lower roller to cause an electric field of 1.0 MV/m (i.e. approaching the breakdown field of air at atmospheric pressure which is ~ 3 MV/m) on each side of the belt.

(a) What is the charging current delivered to the high voltage terminal in µA? Suppose we would like to accelerate the protons to an energy of 3 MeV. Take the radius of the spherical high voltage terminal to be the such that the electric field at the surface of the sphere just below the breakdown field of air

(b) How long does it take to charge the high voltage terminal of the Van de Graaff from zero volts to 3 MV? The beam of 3 MeV protons is focused onto a lithium target. The beam is equivalent to a current of 5 µA.

(c) At what rate do protons strike the target?

(d) At what rate is energy (heat) produced in the target?

(e) Would you consider the Van de Graaff a ‘source of emf ? Why or why not?

Answer 1

given width of the belt, w = 30 cm

speed, v = 20 m/s

electric field near belt, E = 1 MV/m

hence

considering infinite plate approximation

E = sigma/epsilon

hence

10^6 = sigma/epsilon

sigma = 8.86163*10^-6 C/m^2 ( charge density of the belt)

a. now, given that the protons need to accelerate to 3 MeV

hence acceleration potential = 3*10^6 V

radius of the sphere = r

hence, for charge Q on the sphere

kQ/r = 3*10^6

hence

current deliverd = sigma*v*w = 53.169 u A

b. now, kQ/r^2 = 3*10^6

kQ/r^2 = 3*10^6

hence

r = 1 m

and

Q = 3.34*10^-4 C

hence time taken = Q/i = 6.283 s

c. i = 5 uA = q*r ( where r is rate of protons striking the target)

hence

r = 3125*10^10 protons per second

d. rate of heat production = 3 MeV * r = 15 W